How $m^4+4 = 0 \Rightarrow m = 1 \pm i,-1\pm i$? This is given in my module as a part of a problem's solution:
$$m^4 + 4 = 0 $$ $$\Rightarrow m = 1 \pm i,-1\pm i$$
I am not getting how this conversion is taking place,could somebody explain?
 A: One way to look at it is like this:
$$
m^4+4=0\implies m^4=-4\implies m^2=\pm 2i.
$$
But
$$
m^2=2i\implies m=\pm\sqrt{2}\sqrt{i}.
$$
Also,
$$
m^2=-2i\implies m=\pm i\sqrt{2}\sqrt{i}.
$$
Now you can use that fact that there are two square roots of $i$, 
$$
\frac{1+i}{\sqrt{2}}\ \text{and}\ \frac{1+i}{-\sqrt{2}}
$$ 
to simplify each possibility, and get the $4$ values of $m$ you mentioned.
A: Here $m = (-4)^{1/4}$.
Now $-4 = 4(\cos(\theta) + i\sin(\theta))$ in polar form where $\theta = (2n-1)\pi$.
By de Moivre's theorem
$m = (-4)^{1/4} = 4^{1/4}(\cos(\frac{\theta}{4}) + i\sin(\frac{\theta}{4}))$
For $n = 0$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$,
For $n = 1$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$,
For $n = 2$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = -1 + i$,
For $n = 3$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$
A: You can also factor $x^4+4=(x^2-2x+2)(x^2+2x+2)$.
A: HINT $\rm\ -4 = m^4\ \Rightarrow\ \pm 2\ i = m^2 = (a+b\ i)^2 = a^2-b^2 + 2\:a\:b\ i\ \Rightarrow\ |a| = |b| = 1\:$.
